How to find the logarithm of Pauli Matrix? [closed]

Question

When I solve some physics problem, it helps a lot if I can find the logarithm of Pauli matrix.

e.g. $\sigma_{x}=\left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$, find the matrix $A$ such that $e^{A}=\sigma_{x}$.

At first, I find a formula only for real matrix:

$$\exp\left[\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\right]=\frac{e^{\frac{a+d}{2}}}{\triangle}\left(\begin{array}{cc} \triangle \cosh(\frac{\triangle}{2})+(a-d)\sinh(\frac{\triangle}{2}) & 2b\cdot \sinh(\frac{\triangle}{2})\\ 2c\cdot \sinh(\frac{\triangle}{2}) & \triangle \cosh(\frac{\triangle}{2})+(d-a)\sinh(\frac{\triangle}{2}) \end{array}\right)$$

where $\triangle=\sqrt{\left(a-d\right)^{2}+4bc}$

but there is no solution for the formula on this example;

After that, I try to Taylor expand the logarithm of $\sigma_{x}$:

$$ \log\left[I+\left(\sigma_{x}-I\right)\right]=\left(\sigma_{x}-I\right)-\frac{\left(\sigma_{x}-I\right)^{2}}{2}+\frac{\left(\sigma_{x}-I\right)^{3}}{3}... $$

$$ \left(\sigma_{x}-I\right)=\left(\begin{array}{cc} -1 & 1\\ 1 & 1 \end{array}\right)\left(\begin{array}{cc} -2 & 0\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} -\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right) $$

\begin{eqnarray*} \log\left[I+\left(\sigma_{x}-I\right)\right] & = & \left(\begin{array}{cc} -1 & 1\\ 1 & 1 \end{array}\right)\left[\left(\begin{array}{cc} -2 & 0\\ 0 & 0 \end{array}\right)-\left(\begin{array}{cc} \frac{\left(-2\right)^{2}}{2} & 0\\ 0 & 0 \end{array}\right)...\right]\left(\begin{array}{cc} -\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right)\\ & = & \left(\begin{array}{cc} -1 & 1\\ 1 & 1 \end{array}\right)\left(\begin{array}{cc} -\infty & 0\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} -\frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right) \end{eqnarray*}

this method also can't give me a solution.

Answer 1

Observe that \begin{equation} \sigma_{z} = \begin{pmatrix}1&0\\0&-1\end{pmatrix} = \exp(B) = \sum_{r=0}^{\infty} \frac{B^{r}}{r!} \end{equation} with \begin{equation} B = i\pi\begin{pmatrix}2m&0\\0&2n+1\end{pmatrix}, \end{equation} where $m,n\in\mathbb{Z}$.

Next, notice that \begin{equation} \sigma_{x} = U \sigma_{z} U^{\dagger} \end{equation} with \begin{equation} U = \exp(-i\pi\sigma_{y}/4)= \frac{1}{\sqrt{2}}(I - i\sigma_{y})= \frac{1}{\sqrt{2}} \begin{pmatrix}1&-1\\1&1\end{pmatrix}. \end{equation}

Hence, we have \begin{equation} \sigma_{x} = \sum_{r=0}^{\infty} U\frac{B^{r}}{r!} U^{\dagger} = \sum_{r=0}^{\infty} \frac{(UBU^{\dagger})^{r}}{r!} = \exp(A) \end{equation} with \begin{equation} \begin{split} A &= UBU^{\dagger} = i\pi U\left[\left(m+n+\frac{1}{2}\right)I + \left(m-n-\frac{1}{2}\right)\sigma_{z}\right]U^{\dagger}\\ &= i\pi \left[\left(m+n+\frac{1}{2}\right)I + \left(m-n-\frac{1}{2}\right)\sigma_{x}\right]\\ &= i\pi\begin{pmatrix}m + n + 1/2&m - n - 1/2\\m - n - 1/2&m + n + 1/2\end{pmatrix}. \end{split} \end{equation}

Answer 2

As March comments:

$e^{ia(\hat{n}\cdot\vec{\sigma})}=I\cos(a)+i(\hat{n}\cdot\vec{\sigma})\sin(a)$ would be of use

For this example, set $a=\frac{\pi}{2}$, $\hat{n}\cdot\vec{\sigma}=\sigma_{x}$,

the Euler's formula is rewritten as:

$$ e^{i\frac{\pi}{2}\sigma_{x}}=i\sigma_{x}=e^{i\frac{\pi}{2}I}\sigma_{x} \tag{1}$$

Since $[\sigma_{x},I]=0$, we can combine two exponentials to get:

$$ e^{i\frac{\pi}{2}\sigma_{x}-i\frac{\pi}{2}I}=\sigma_{x} \tag{2}$$

Finally, we get a solution for $A$: $A=i\frac{\pi}{2}\sigma_{x}-i\frac{\pi}{2}I=\left(\begin{array}{cc} -i\frac{\pi}{2} & i\frac{\pi}{2}\\ i\frac{\pi}{2} & -i\frac{\pi}{2} \end{array}\right)$.

If we consider periodic conditions in equations (1) and (2), we may get the same result as higgsss gets here.

The procedure is identical for other Pauli Matrices except the subscript, as a result:

$A=i\frac{\pi}{2}(\sigma_{j}-I)$ is a solution for $e^{A}=\sigma_{j}$,$j\in\{x,y,z\}$.

Answer 3

As Ruslan comments:

... You can find it if you switch to eigenbasis of the Pauli matrix, compute the logarithms of its diagonal elements, then switch back.

and, since a Pauli matrix is a normal operator (commutes with its Hermitian conjugate), it can always be diagonalized by a unitary matrix of eigenvalues, hence Ruslan's suggested method is failsafe here. Note, however, that the logarithm has branches and it is not unique.

Further to Ruslan's suggestion, there is another neat trick that applies to $SU(2)$ and $SO(3)$ and the three Pauli matrices, multiplied by $i$, all belong to $SU(2)$. This is the method used in [1] to find a closed form expression for the Campbell Baker Hausdorff series for these groups.

From the characteristic equation $\lambdaˆ2 + rˆ2 =0$ for a general superposition $H=i\,r_x \,\sigma_x+i\,r_y\,\sigma_y + i\,r_z\, \sigma_z$ belonging to the Lie algebra $\mathfrak{su}(2)$ (here $r=\sqrt{r_xˆ2+r_yˆ2+r_zˆ2}$) we can prove (from the Taylor series) that:

$$\exp(H) = \cos(r) \,\mathrm{id} + \frac{\sin(r)}{r}\,H\tag{1}$$

Your Pauli matrix times $i$ $\sigma$ is the $\exp(H)$ here and we wish to find $H$ such that $i\,\sigma=eˆH$. Note from (1) that we almost get the logarithm if we take the skew-Hermitian part of $\sigma$: the part $ \cos(r) \,\mathrm{id}$ is Hermitian. So we use the unique decomposition of any matrix into its Hermitian and skew Hermitian parts to find:

$$\frac{\sin(r)}{r}\,H = \frac{1}{2}((i\,\sigma)-(i\,\sigma)ˆ\dagger)= \frac{i}{2}(\sigma+\sigmaˆ\dagger)\tag{2}$$

and we're almost there. Now we simply need to find out what $r$ is. Since we can always chose the sign of $H$, we can assume that $r\geq0$, whence we can find our result simply by taking the Frobenius norm of both sides of (2):

$$\frac{\sin (r)}{r}\, \|H\|= \|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|\tag{3}$$

Now $H/r$ is simply the normalized (unit Frobenius norm version of) $H$, therefore:

$$\sin(r) = \|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|\tag{4}$$

whence we now have our formula for the logarithm (or, more precisely, the principal branch thereof, since the logarithm has branches):

$$\log (i\,\sigma) = \frac{\arcsin\|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|}{\|\frac{1}{2}(\sigma+\sigmaˆ\dagger)\|}\;\frac{i}{2}(\sigma+\sigmaˆ\dagger)\tag{5}$$

and then you simply need to take account of the factor of $i$ to find $\log\sigma$ given $\log(i\,\sigma)$.

[1] K. Engø, "On the BCH-Formula in SO(3)", BIT Numerical Mathematics 41 (2001), no.3, pp629--632.